Test: Algebraic Solutions of Linear Inequalities in One Variable - Grade 12 MCQ

x>7 and 7>4 => x>7>4 => x>4.
If x>7 then x cannot be less than 6.
If x=11 then x>7 and x>9.
If x=11 then x>7 and x<14.

If we multiply by negative number on both sides of inequality then sign of inequality will change i.e. if x>7 then (-1) x < (-1)7 => -x<-7.

20x<100
Dividing by 20 on both sides, x< (100/20) => x<5
Since x is a positive integer so x = 1,2,3,4.

20x≤100
Dividing by 20 on both sides, x ≤ (100/20) => x≤5
Since x is a natural number so x = 1,2,3,4,5.

10x≤50
Dividing by 10 on both sides, x ≤ (50/10) => x≤5
Since x is a whole number so x = 0,1,2,3,4,5.

2x+1>5
=>2x>5-1
=>2x>4 => x>2.

x-1>-x+7
=>2x>8 => x>4.

Average is at least 30 marks.
Let x be the marks in 3^rd test.
Average = (20+25+x)/3 ≥30
=>45+x≥90 => x≥90-45 => x≥45.
Minimum marks in 3^rd test should be 45.

Let two numbers be x and x+2.
x + x+2 >10 => 2x>8 => x>4
and x<8
and x+2<8 => x<6.
4<x<6 => x can be 5.
For x =5, x+2=7
So, Pairs of odd consecutive positive integers are (5,7).

Let shortest side be x. Then longest side = 2x.
Third side = 2x - 4.
Given, perimeter of triangle is at least 61 cm
=>x + 2x + 2x - 4 ≥ 61 => 5x ≥ 65 = x ≥ 13.
Minimum length of the shortest side is 13 cm.